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Let $F:{\mathbb R}^n \to {\mathbb R}^n$ have coordinate functions in $L^2$. Suppose $F$ is weakly divergence-free, that is, $\int_{{\mathbb R}^n} F \cdot \nabla \varphi = 0$ for all $\varphi \in C_0^\infty({\mathbb R}^n)$. Must $F$ be continuous?

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No, but $F$ must be continuous across any $(n-1)$-hypersurface along its normal direction. –  Shuhao Cao Dec 2 '11 at 16:18
    
@MathChief: Can you recommend a counterexample and a reference for your assertion? –  Stefan Smith Dec 2 '11 at 16:45
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up vote 1 down vote accepted

Consider a vector field defined on a square in $\mathbb{R}^2$, on the upper plane, define $F = \left(\begin{matrix} y-1\\x \end{matrix} \right)$ while on the lower plane define $F = \left(\begin{matrix} 1-y\\x \end{matrix} \right)$, you could check $F$ is weakly divergence free, and $F\cdot \vec{n}$ is continuous across the $x$-axis, where $\vec{n} = \left(\begin{matrix} 0\\1 \end{matrix} \right)$ is the normal vector to the $x$-axis.

However, $\displaystyle \lim_{y\to 0^+} F\times \vec{n} = -\lim_{y\to 0^-} F\times \vec{n}$, which basically says $F$ is not continuous along $x$-axis.

The idea behind this is: pick any domain $\Omega \subset \mathbb{R}^n$, on any $(n-1)$-hypersurface $S$ within the domain which cuts it into two parts: $\Omega_1$ and $\Omega_2$, integrate by parts separately we have $$ 0 = \int_{\Omega} F\cdot \nabla \phi = \int_{\Omega_1} \phi\,\mathrm{div} F + \int_{\Omega_2} \phi\,\mathrm{div} F + \int_{S} (F\cdot \vec{n}_1+F\cdot \vec{n}_2)\phi\,ds + \int_{\partial \Omega} F\cdot \vec{n}\,\phi\,ds $$ For the integration by parts formula to hold, we must have $F\cdot \vec{n}_1+F\cdot \vec{n}_2=0$,ie, normal direction continuity, but for a divergence integrable field you can't really tell how its tangential trace $F\times \vec{n}$ behaves on the hypersurfaces.

You could learn more of this by Googling the keywords: tangential trace for $H(\mathrm{div})$ and $H(\mathbf{curl})$, normally the analysis is done in 3D setting.

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for your counterexample, $F \cdot <1, -1>$ is not continuous alone the line $y=x$. –  Stefan Smith Dec 7 '11 at 14:26
    
I am sorry that I phrased my argument a bit ambiguous, in order that the integration by parts formula to hold, $F\cdot \vec{n}$ is continuous across the hypersurface almost everywhere, which means it can be discontinuous upto a measure zero set of the $(n-1)$-dimensional measure you are integrating to. –  Shuhao Cao Dec 8 '11 at 0:48
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